7 Cauchy’s theorem
Definition 7.1 Let \(D\subset\C\) and \(\ga_0, \ga_1\colon[a,b]\to D\) curves in \(D.\)
A homotopy in \(D\) between paths \(\ga_0, \ga_1\) is a continuous map \[\Ga\colon[0,1]\t[a,b]\longra D, (s,t)\longmapsto\Ga_s(t),\] such that \(\Ga_0(t)=\ga_0(t)\) and \(\Ga_1(t)=\ga_1(t)\) for all \(t\in[a,b].\) Then \(\ga_0, \ga_1\) are called (freely) homotopic paths in \(D.\)
If, additionally, \(\Ga_s(a)=p\) and \(\Ga_s(b)=q\) are constant in \(s\in[0,1],\) we call \(\Ga\) a path homotopy in \(D\) and \(\ga_0, \ga_1\) path-homotopic in \(D\).
If \(\ga_0, \ga_1\) are loops, a homotopy of loops in \(D\) is a homotopy \(\Ga\) in \(D\) with the additional property that \(\Ga_s\) is a loop for each \(s\in[0,1].\) Then \(\ga_0, \ga_1\) are called (freely) homotopic loops in \(D.\)
A loop is null-homotopic in \(D\) if there is a homotopy of loops in \(D\) to the constant loop.
Remark 7.1. In the following we will also suppose that \(\Ga\) is piecewise C1, meaning there exist subdivisions \[0=s_0<s_1<\cdots<s_m=1,\qquad a=t_0<t_1<\cdots<t_n=b\]
such that each restriction \(\Ga|_{[s_{j-1},s_j]\t[t_{k-1},t_k]}\) is continuously differentiable.
Example 7.1
A set \(D\) is star-shaped if there exists a focal point \(z_0\in D\) such that for each \(z\in D\) the straight line segment \(tz+(1-t)z_0,\) \(t\in[0,1],\) is contained in \(D.\) Disks, the complex plane, and rectangles are star-shaped. Every loop \(\ga\) in a star-shaped domain is null-homotopic. If the focal point agrees with the base of the loop, the path homotopy is
\[\Ga_s(t) = (1-t)\ga(t) + tz_0. \tag{7.1}\]
In general, the path homotopy is more complicated to write down. We omit it, since we will not need this fact below.
Theorem 7.1 (Cauchy’s theorem) Let \(f\colon U\to \C\) be a holomorphic function on an open set. Let \(\ga_0, \ga_1\) be piecewise C1 curves in \(U\) that are path-homotopic in \(U.\) Then
\[\int_{\ga_0}f(z)dz = \int_{\ga_1}f(z)dz.\]
Proof.
Pick a path homotopy \(\Ga\colon[0,1]\t[a,b]\to U.\) By reparameterizing we may assume \([a,b]=[0,1].\) Write \(R^{(0)}=[0,1]\t[0,1]\) for the domain of \(\Ga\) and let \(\6 R^{(0)}\) be its piecewise linear boundary path, with the obvious counterclockwise parameterization (see Figure 7.2).
The boundary path has four segments and the path integral along each constant path vanishes. Using Proposition 6.2(b),(c), we then find that \[\int_{\ga_0} f(z)dz-\int_{\ga_1} f(z)dz=\int_{\Ga(\6 R^{(0)})} f(z)dz.\]
Subdivide \(R^{(0)}\) into four congruent rectangles \(R^{(0)}_i\) as in Figure 7.3. Using the fact that the path integrals in opposite directions cancel, we find \[\left|\int_{\Ga(\6 R^{(0)})} f(z)dz\right| = \left|\sum_{i=1}^4 \int_{\Ga(\6 R^{(0)}_i)} f(z)dz\right|\leqslant \sum_{i=1}^4 \left|\int_{\Ga(\6 R^{(0)}_i)} f(z)dz\right|.\]
Let \(R^{(1)}\) be the rectangle \(R_i^{(0)}\) for which \(\left|\int_{\Ga(\6 R^{(0)}_i)} f(z)dz\right|\) is maximal. Then \[\left|\int_{\Ga(\6 R^{(0)})} f(z)dz\right|\leqslant4\left|\int_{\Ga(\6 R^{(1)})} f(z)dz\right|.\]
Now repeat this process with \(R^{(1)}\) to obtain \(R^{(2)}\) and so forth. This yields a sequence of rectangles \(R^{(n)}\) as in Figure 7.4 with sides of length \(2^{-n}\) and \[\left|\int_{\Ga(\6 R^{(0)})} f(z)dz\right|\leqslant4^n\left|\int_{\Ga(\6 R^{(n)})} f(z)dz\right|. \tag{7.2}\]
As the side lengths tend to zero, the midpoints \((s^{(n)},t^{(n)})\) of the rectangles \(R^{(n)}\) are a Cauchy sequence, so converge to some limit \((s_0,t_0).\) Let \(z_0=\Ga_{s_0}(t_0).\) Since \(f\) is complex differentiable at \(z_0,\) we may write \[f(z)=f(z_0) + (z-z_0)f'(z_0) + (z-z_0)\rho(z),\qquad\lim_{z\to z_0}\rho(z)=0. \tag{7.3}\] By Example 6.4,\[\int_{\Ga(\6 R^{(n)})}\left(f(z_0) + (z-z_0)f'(z_0)\right)dz=0. \tag{7.4}\]
As \(\Ga\) is piecewise C1, its derivative is bounded in norm by some \(C>0.\) Let \(\ep>0.\) Pick \(n_0\in\N\) such that for all \(n\geqslant n_0\) and \(z\in R^{(n)}\) we have \(|\rho(z)|<\ep.\) As the side lengths of \(R^{(n)}\) are \(2^{-n}\) and \(z_0\in R^{(n)},\) we have \(|z-z_0|\leqslant\sqrt{2}2^{-n}\) for all \(z\in\6 R^{(n)}.\) Combining this with Equation 7.3 and Equation 7.4, we can estimate
\[\begin{align*} \left|\int_{\Ga(\6 R^{(n)})} f(z)dz\right|&= \left|\int_{\Ga(\6 R^{(n)})}(z-z_0)\rho(z)dz\right|\\ &\leqslant L(\Ga(\6 R^{(n)}))\sqrt{2}2^{-n}\ep = 4C\cdot 2^{-n}\sqrt{2}2^{-n}\ep \end{align*}\]
The last inequality follows by Equation 6.7.
Hence \[\left|\int_\ga f(z)dz\right|=\left|\int_{\Ga(\6 R^{(0)})} f(z)dz\right|\overset{}{\leqslant}4C\sqrt{2}\ep.\] The final inequality follows from Equation 7.2.
As \(\ep>0\) is arbitrary, the left hand side must be zero.
Theorem 7.2 Let \(\ga_0, \ga_1\) be piecewise C1 loops in \(U\) that are freely homotopic in \(U.\) Suppose \(f\colon U\to\C\) is a holomorphic function. Then
\[\int_{\ga_0}f(z)dz = \int_{\ga_1}f(z)dz.\]In particular, if \(\ga\) is a loop that is homotopic in \(U\) to the constant loop, then\[\int_\ga f(z)dz=0. \tag{7.5}\]
Proof.
Let \(\Ga\) be the homotopy between \(\ga_0\) and \(\ga_1.\) Let \(\eta(s)=\Ga_s(0).\) Figure 7.5 describes the construction of a path homotopy between \(\eta\ast\ga_0\ast(-\eta)\) and \(\ga_1.\)
Hence Theorem 7.1 implies our claim, using the fact that the path integrals over \(\eta\) and \(-\eta\) cancel each other.
Questions for further discussion
- How should Figure 7.5 be interpreted? Can you give a formula for the path homotopy (consider cases)?
Give a counterexample to Cauchy’s theorem when (i) \(\ga\) is not null-homotopic (ii) \(f(z)\) is not holomorphic
Does Cauchy’s theorem hold for \(f(z)=\ol{z}\)?
7.1 Exercises
Sketch the curves \[\begin{align*} \ga_0\colon[-1,1]&\longra\C,&\ga_0(t)&=t,\\ \ga_1\colon[-1,1]&\longra\C,&\ga_1(t)&=e^{i\pi\frac{1-t}{2}} \end{align*}\] and show that they are path-homotopic.
- Let \(0<r_0<r_1\) and \(z_0\in\C.\) Prove that the loops \(\partial D_{r_0}(z_0)\) and \(\partial D_{r_1}(z_0)\) are freely homotopic in the closed annulus \(\ol A_{r_0,r_1}(z_0).\)
Let \(\al+i\be\in\C.\) Determine \[\int_a^b e^{(\al+i\be)t} dt\] to compute \[\int_a^b e^{\al t}\cos(\be t)dt.\]
Let \(\C^-=\C\setminus(-\iy,0]\) be the slit plane.
- Show that any two points in \(\C^-\) may be connected by a path in \(\C^-.\) Hence \(\C^-\) is path-connected.
- Show that every closed curve \(\ga\colon[a,b]\to\C^-\) is null-homotopic in \(\C^-\) by finding a homotopy \[\Ga\colon[0,1]\t[a,b]\longra\C^-, (s,t)\longmapsto\Ga_s(t)\] satisfying \(\Ga_0(t)=\ga(t)\) and \(\Ga_1(t)=1\) for all \(t\in[a,b].\) Hence \(\C^-\) is simply-connected.
- Prove the analogues of a. and b. for a disk \(D_r(z_0).\)
- Use Cauchy’s Theorem to prove that the punctured plane \(\C^\t=\C\setminus\{0\}\) is not simply-connected. That is, there exists a closed curve in \(\C^\t\) that is not null-homotopic in \(\C^\t.\)
Let \(0<b<1.\)
- Using the geometric series, find the power series expansion \[\frac{1}{z-1/b}=\sum_{n=0}^\iy a_n(z-b)^n\] with center \(z_0=b\) and determine the radius of convergence \(\rho.\)
- Use a. to show that for all \(0<r<\rho\) \[\int_{\partial D_r(b)}\frac{dz}{(z-b)(z-1/b)}=\frac{2\pi i}{b-1/b}.\]
- Use c. to compute \[\int_0^{2\pi}\frac{dt}{1-2b\cos(t)+b^2}.\]
Let \(\ga\colon[0,1]\to D\) be a curve in \(D\subset\C\) and let \(-\ga\colon[0,1]\to D,\) \((-\ga)(t)=\ga(1-t)\) be the opposite curve. Prove that \(\ga\ast(-\ga)\) is path-homotopic to a constant loop.